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10 Mar 2021, 07:00
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:31) correct
15%
(01:56)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
In March 2010, my wife Sarmishtha was eight times as old as my daughter Surichi. In March 2020, she will be three times as old as my daughter. How old was my daughter in March 2010 ?
(A) 4 years
(B) 6 years
(C) 18 years
(D) 24 years
(E) 26 years
(A) 4 years
(B) 6 years
(C) 18 years
(D) 24 years
(E) 26 years
10 Mar 2021, 07:33
Kudos
Bookmarks
Bunuel
In March 2010, my wife Sarmishtha was eight times as old as my daughter Surichi.
Let x = Surichi's age in 2010
So, 8x = Sarmishtha's age in 2010
This also means...
x + 10 = Surichi's age in 2020
8x + 10 = Sarmishtha's age in 2020
In March 2020, she will be three times as old as my daughter.
We can write: (Sarmishtha's age in 2020) = 3(Surichi's age in 2020)
Substitute to get: 8x + 10 = 3(x + 10)
Expand: 8x + 10 = 3x + 30
Subtract 3x from both sides: 5x + 10 = 30
Subtract 10 from both sides: 5x = 20
Solve: x = 20/5 = 4
Answer: A
General Discussion
Archit3110
Major Poster
10 Mar 2021, 07:05
Kudos
Bookmarks
Bunuel
let Wife be denoted by W and Daughter by D
in 2010
W=8D---(1)
and 10 years later 2020
W+10 = 3*(D+10)
substitute from (1)
8D+10=3D+30
5D= 20
D= 4
option A
